91Ó°ÊÓ

The null hypothesis states that the population mean is equal to the value mentioned in the claim. If the null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis.

\(\begin{array}{l}{H_0}:p = 0\\{H_a}:p \ne 0\end{array}\)

The formula for the value of the test statistic is given by,\(z = \frac{{\hat p - {p_0}}}{{\sqrt {\frac{{{p_0}\left( {1 - {p_0}} \right)}}{n}} }}\).

The sample proportion is calculated by dividing the number of successes by the sample size. \(\hat p = \frac{x}{n}\)

(a)

Let the appropriate hypotheses be\({H_0}:p = 0.05\)versus\({H_0}:p \ne 0.05\).

Let the given be:

\(\begin{array}{c}{p_0} = 0.4\\n = 150\end{array}\)

Sample proportion:

\(\begin{array}{c}\hat p = \frac{{40}}{{500}}\\ = 0.08\end{array}\)

The value of the test-statistic:

\(\begin{array}{c}z = \frac{{0.08 - 0.05}}{{\sqrt {0.05(1 - 0.05)/500} }}\\ = 3.07\end{array}\)

The P-value of the alternative hypothesis\({H_a}:p \ne {p_0}\)is,

\(\begin{array}{l} = 2 \times (Area under the standard normal curve to the right of|z|)\\ = 2 \times P(z \ge 3.07)\\ = 2 \times 0.0011\\ = 0.0022\end{array}\)

Hence, the null hypothesis has been rejected because \(0.0022 < 0.01\). The company’s premise is not correct. The same conclusions would be the same for \(\alpha = 0.05\).

(b)

The type II error in case when the alternative hypothesis\({H_a}:p \ne {p_0}\) is

\(\beta \left( {{p^\prime }} \right) = \Phi \left( {\frac{{{p_0} - {p^\prime } + {z_{\alpha /2}} \cdot \sqrt {{p_0} \cdot \left( {1 - {p_0}} \right)/n} }}{{\sqrt {{p^\prime } \cdot \left( {1 - {p^\prime }} \right)/n} }}} \right) - \Phi \left( {\frac{{{p_0} - {p^\prime } - {z_{\alpha /2}} \cdot \sqrt {{p_0} \cdot \left( {1 - {p_0}} \right)/n} }}{{\sqrt {{p^\prime } \cdot \left( {1 - {p^\prime }} \right)/n} }}} \right)\)

Hence, in case of\(10\% = 0.1\) of all current customers qualify, the type II error (required in the exercise) is

\(\begin{array}{c}\beta (0.1) = \Phi \left( {\frac{{0.05 - 0.1 + 2.58 \cdot \sqrt {0.05 \cdot (1 - 0.05)/500} }}{{\sqrt {0.1 \cdot (1 - 0.1)/500} }}} \right) - \Phi \left( {\frac{{0.05 - 0.1 - 2.58 \cdot \sqrt {0.05 \cdot (1 - 0.05)/500} }}{{\sqrt {0.1 \cdot (1 - 0.1)/500} }}} \right)\\ \approx \Phi ( - 1.85) - 0\\ = 0.0332\end{array}\)